#### A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by and for on m/s. It repeats this cycle till it reaches the height of 20 m.a) At what time is its velocity maximum?b) At what time is its average velocity maximum?c) At what times is its acceleration maximum in magnitude?d) How many cycles are required to reach the top?

(a) for the velocity to be maximum,

Thus,

(b)Now, average velocity

Let us integrate both the sides from 0 to 3

Thus,

Now, average velocity

$t=2.36s$

Thus, the average velocity is maximum at 2.36 seconds.

(c) When the body returns at its mean position or changes direction in periodic motion, time for acceleration is maximum.

Thus, the acceleration is maximum at t = 3 sec.

(d)Now, for 3 to 6 sec

Integrate from 3 to 6 s

.............because distance is in downward direction

Thus, net distance

Thus, in three cycle

Remaining height will be

The monkey can climb up to 9m without slipping but in the 4th cycle it will slip and the height remaining to climb will be 6.5 m.

Net no. of cycle = 4.